How would you prove algebraically that the following function is one to one? f(x)= (x+3)^2 , x>= -3?
So, do you mean what is 5x+3=4x? If you do, I don't think that's algebraically possible, but I'm not sure. You have to have at least one more number without a variable on the 4x side for the equation to be solvable.
3x-2 = 2x+3 3x-2x = 3+2 x = 5
x+3=14 x+3-3 = 14-3 (subtract 3 from both sides to solve for x) x = 11
The variable is a. You can solve for it, like this: 8a - 9 = 7a - 6 8a = 7a + 3 8a - 7a = 3 a = 3
34 = 81 so x is 4
8 1/3 = ab^-1, 1.8 =ab^2
Expressed algebraically, this is equal to 3x + 8=y.
3
Well if you mean you have to solve w^3+27 then I guess make it = 0 then solve it w^3+27=0 -27 w^3=27 3rd root w=-3 ~¤~Amy~¤~
y - 3 = 0 y = 3
m= -3
3
3
-2 plus 5 equals +3
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
10 - 3 217 equals - 3 207.